If sin (A + B) = 1 and cos (A − B) = 1,

Given:

$\sin (A+B)=1$...(1)

$\cos (A-B)=1$....(2)

We know that,

$\sin 90^{\circ}=1$.....(3)

$\cos 0^{\circ}=1$.....(4)

Now by comparing equation (1) and (3)

We get,

$A+B=90 \ldots \ldots(5)$

Now by comparing equation (2) and (4)

We get,

$A-B=0 \ldots \ldots(6)$

Now to get the values of A and B, let us solve equation (5) and (6) simultaneously

Therefore by adding equation (5) and (6)

We get,

Therefore,

$2 A=90$

$\Rightarrow A=\frac{90}{2}$

$\Rightarrow A=45^{\circ}$

Hence $A=45^{\circ}$

Now by subtracting equation (6) from equation (5)

We get,

Therefore,

$2 B=90$

$\Rightarrow B=\frac{90}{2}$

$\Rightarrow B=45^{\circ}$

Hence $B=45^{\circ}$

Therefore the values of A and B are as follows

$A=45^{\circ}$ and $B=45^{\circ}$